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%\newcommand{\showsolution}{2} %%设置showsolution=2, 编译生成试卷解答与评分标准、阅卷与试卷袋归档

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%% 填写课程信息：
\newcommand{\CourseName}{复变函数}
\newcommand{\CourseNumber}{162250220}
\newcommand{\CourseStudents}{2022级数学与应用数学}
\newcommand{\CourseTerm}{2024 $\sim$ 2025 学年 第 一 学期}
\newcommand{\ExamAB}{A}
\newcommand{\ExamContents}{本次考试的主要内容是：
复数的代数运算、
%用复数方程表示平面几何图形、
解析函数的概念、
解析函数与二元向量值函数的联系、
计算有理函数的部分分式、
计算三角函数、
使用交比找出符合条件的线性变换、
解析函数的保角性质、
道路积分与路径的关系、
计算复值函数的道路积分、
可去奇点的充分必要条件、
使用留数定理计算实积分、
使用幅角原理计算解析函数的零点的个数。
%将亚纯函数展开成幂级数。
%考试内容覆盖教学大纲的大部分重要内容。
}

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\usepackage{titling}
\setlength{\droptitle}{-2cm}   % 标题上移2cm

\author{王立庆（\CourseStudents）}
\title{\CourseName 考试 \ExamAB 解答 }
\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
\date{2024年12月19日}

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\begin{document}

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\begin{center}
\includegraphics [width=0.85\textwidth, height=4.1cm]{lixin-pan-new.eps}
\end{center}

\vspace{-0.5cm}

\begin{center}
{\Large \bf 上海立信会计金融学院期终考试卷  \hspace{0.3cm} \underline{\,\ExamAB\, } 卷 }

\vspace{0.3cm}

{\large \CourseTerm }

\vspace{0.3cm}

{\large \bf \underline{\,\CourseStudents\,} 《\underline{\,\CourseName\,}》 课程代码：\underline{\,\CourseNumber\,} }

\vspace{0.3cm}

（本场考试属\underline{ \, 闭 \, }卷考试，考试时间共\underline{ \, 90 \,  }分钟，不准使用计算器）共\underline{ \, \pageref{LastPage} \, }页 

\vspace{0.7cm}

班级 \underline{\hspace{3.5cm}} 学号 \underline{\hspace{3.5cm}} 姓名 \underline{\hspace{3.5cm}} 

\end{center}

\vspace{-0.2cm}

\begin{table}[h]
\centering
\renewcommand{\arraystretch}{1.2}
\begin{tabular}{
|>{\centering\arraybackslash}p{0.8cm}|>{\centering\arraybackslash}p{0.7cm}|>{\centering\arraybackslash}p{0.7cm}|
>{\centering\arraybackslash}p{0.7cm}|>{\centering\arraybackslash}p{0.7cm}|>{\centering\arraybackslash}p{0.7cm}|
>{\centering\arraybackslash}p{0.7cm}|>{\centering\arraybackslash}p{0.7cm}|>{\centering\arraybackslash}p{0.7cm}|
>{\centering\arraybackslash}p{0.7cm}|>{\centering\arraybackslash}p{0.7cm}
|>{\centering\arraybackslash}p{1.0cm}|>{\centering\arraybackslash}p{1.2cm}|>{\centering\arraybackslash}p{1.2cm}|}
\hline
题号 &一&二&三&四&五&六&七&八&九&十&总分&合成人签名&审核人签名 \\
\hline
得分 $\,\,\,\,\,\,\,\,$ &&&&&&&&&&&&& \\
\hline
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\end{table}

\vspace{0.2cm}

本次考试共10题，每题10分。

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\begin{center}

{\Large \bf 上海立信会计金融学院期终考试卷 \,\, \underline{\, \ExamAB \, } 卷\,\, 解答与评分标准}

\vspace{0.3cm}

{\large \CourseTerm }

\vspace{0.3cm}

{\large \bf \underline{ \, \CourseStudents \, } 《\underline{ \, \CourseName \, }》 课程代码：\underline{ \, \CourseNumber \, }  }

\end{center}

\vspace{0.2cm}

本次考试共10题，每题10分。

\vspace{0.2cm}

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\begin{enumerate}

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%\newpage 
\item %1
%\begin{enumerate}
%\item 
Suppose that $|a| < 1$ and $|b| < 1$. Prove 
$$\left\vert \frac{a-b}{1-\bar{a}b} \right\vert < 1. $$
%\item  Let $f(z)=u+iv$ be an analytic function. Shows that $|f'(z)|^2$ is the Jacobian of $u$ and $v$ with respect to $x$ and $y$.
%\end{enumerate}

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{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  Since $|z|^2=z\bar{z}$, we have 
$$\left\vert \frac{a-b}{1-\bar{a}b} \right\vert^2 
=  \frac{(a-b)(\bar{a}-\bar{b})}{(1-\bar{a}b)(1-a\bar{b})}
=\frac{a\bar{a} + b\bar{b} -a\bar{b} - \bar{a}b}{1-\bar{a}b - a\bar{b} + a\bar{a}b\bar{b}}.
$$
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\item  Both the denominator and the numerator are non-negative real numbers, their difference is 
\begin{equation*}
\begin{aligned}
& (1-\bar{a}b - a\bar{b} + a\bar{a}b\bar{b}) - (a\bar{a} + b\bar{b} -a\bar{b} - \bar{a}b) \\ 
&= 1 + a\bar{a}b\bar{b} - a\bar{a} - b\bar{b} = (1-a\bar{a})(1-b\bar{b}). 
\end{aligned}
\end{equation*}
\ifnum\showsolution=2
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\item  Since $|a| < 1$ and $|b| < 1$, we have $a\bar{a}<1$ and $b\bar{b}<1$. 
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\item  Thus the denominator is greater than the numerator, and the proof is complete. 
\ifnum\showsolution=2
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\fi


\end{enumerate}

}

\fi

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\vspace{0.2cm}

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%\newpage 
\item %2
%When does $az + b\bar{z} + c = 0$ represent a line, a point, and an empty set?
Let $f(z)=u+iv$ be an analytic function. From the definition of analytic functions, show that $|f'(z)|^2$ is the Jacobian of $u$ and $v$ with respect to $x$ and $y$.

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{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  Since $f(z)$ is analytic, the limit $f'(z) = \lim\limits_{h\to 0} \frac{f(z+h)-f(z)}{h}$ exists. 

\item  When we choose real values and purely imaginary values for $h$, we obtain 
\begin{equation*}
\begin{aligned}
f'(z) &= \lim\limits_{\Delta x \to 0} \frac{f(z+\Delta x)-f(z)}{\Delta x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}, \\ 
f'(z) &= \lim\limits_{i\Delta y \to 0} \frac{f(z+i\Delta y)-f(z)}{i\Delta y} = -i\left(\frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y}\right).  
\end{aligned}
\end{equation*}
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\item  Thus we obtain Cauchy-Riemann Equations 
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\,\, \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}. $$
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\item  By the definition of modulus of a complex number, 
$$
|f'(z)|^2 = \left(\frac{\partial u}{\partial x} \right)^2 + \left(\frac{\partial v}{\partial x} \right)^2. 
$$
\ifnum\showsolution=2
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\item  The Jacobian of $u$ and $v$ with respect to $x$ and $y$ is 
$$
\frac{\partial (u,v)}{\partial (x,y)}
=\begin{vmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ 
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \\ 
\end{vmatrix} 
= \frac{\partial u}{\partial x}\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\frac{\partial v}{\partial x}. 
$$
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
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\item  By Cauchy-Riemann Equations, we see these two expressions are equal. 


\end{enumerate}

}
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\fi

\vspace{0.2cm}

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%\newpage 
\item %3
Develop the following expression in partial fractions by looking for the singular part at each pole, 
$$
f(z) = \frac{1}{(z-1)(z-2)^2}. 
$$

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{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  The function $f(z)$ has poles at $z=1$ and $z=2$. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  For $z=1$, we change variables $z = 1+\frac{1}{\zeta}$, and get 
$$f(z) =\frac{1}{\frac{1}{\zeta}(\frac{1}{\zeta}-1)^2} = \frac{\zeta^3}{(\zeta-1)^2} = \zeta + \frac{2\zeta^2-\zeta}{\zeta^2-2\zeta+1}. $$

\item  Thus the singular part of $f(z)$ at $z=1$ is $\zeta = \frac{1}{z-1}$. 
\ifnum\showsolution=2
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\item  For $z=2$, we change variables $z = 2+\frac{1}{\zeta}$, and get 
$$f(z) =\frac{1}{(1+\frac{1}{\zeta})\frac{1}{\zeta^2}} = \frac{\zeta^3}{\zeta+1} = \zeta^2-\zeta +\frac{\zeta}{\zeta+1}. $$

\item  Thus the singular part of $f(z)$ at $z=2$ is $\zeta^2-\zeta = (\frac{1}{z-2})^2-\frac{1}{\zeta-2}$. 
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\item  Collecting the singular parts, we obtain $$\frac{1}{(z-1)(z-2)^2} = \frac{1}{z-1} + \frac{1}{(z-2)^2} - \frac{1}{z-2} + c,$$
where $c$ is a constant. 
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\item  We see $c=0$ by taking $z\to\infty$. 
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\end{enumerate}

}
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\fi

\vspace{0.2cm}

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%\newpage 
\item %4
\begin{enumerate}
\item  Find the value of $\sin i + \cos i$.
\item  Find the linear transformation which carries $0, 1, i$ into $2, 3, 4$.
\end{enumerate}

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{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  By the definitions of sine and cosine in terms of the exponential function, we have 
$$
\sin i + \cos i = \frac{e^{ii}-e^{-ii}}{2i} + \frac{e^{ii}+e^{-ii}}{2} = \frac{1}{2}(e^{-1}+e) + \frac{i}{2}(e-e^{-1}).  
$$
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\item  The linear transformation has the form $(z,z_1,z_2,z_3) = (w,w_1,w_2,w_3)$. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
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\item  We substitute the given numbers, $$(z,0,1,i)=(w,2,3,4). $$

\item  By the definition of cross ratio, we see that $$\frac{z-1}{z-i}: \frac{0-1}{0-i} = \frac{w-3}{w-4}:\frac{2-3}{2-4}. $$
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\item  Simplifying this equation, we find the linear transformation which carries $0, 1, i$ into $2, 3, 4$ is 
$$w= \frac{(4i-6)z+2i}{(i-2)z+i}. $$
\ifnum\showsolution=2
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\end{enumerate}

}
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\fi

\vspace{0.2cm}

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%\newpage 
\item %5
Let $w=f(z)$ be an analytic function in the region $\Omega$. Let $z_0\in \Omega$ and assume $f'(z_0)\neq 0$.  
Prove that two curves which form an angle at $z_0$ are mapped upon curves forming the same angle. 

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{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  Let $z=z(t), \alpha\le t\le \beta$ be a differentiable curve $\gamma$ in the $z$ plane, and $z_0=z(t_0)$. 
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\item  The tangent vector of the curve $\gamma$ at $z_0$ is $z'(t_0)$. 
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\item  The image curve in the $w$ plane is $w=f(z(t)), \alpha\le t\le \beta$. 
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\item  The tangent vector of the image curve at $w_0=f(z_0)$ is $f'(z_0)\cdot z'(t_0)$. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  The directions of these two tangent vectors are related by the following equation,
$$
\mathrm{arg}\, w'(t_0) = \mathrm{arg}\, f'(z_0) + \mathrm{arg}\, z'(t_0), 
$$
thus the angle between these vectors is $\mathrm{arg}\, f'(z_0)$, and it is independent of the curve. 
\ifnum\showsolution=2
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\end{enumerate}

}
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\vspace{0.2cm}

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%\newpage 
\item %6
Let $p(x,y)$ and $q(x,y)$ be continuous functions in a region $\Omega$. Prove that the line integral $$\int_\gamma p dx + q dy,$$ defined in $\Omega$, depends only on the end points of $\gamma$ if and only if there exists a function $U(x,y)$ in $\Omega$ with the partial derivatives $\partial U/\partial x = p$ and $\partial U/\partial y= q.$

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{\color{red} 解答：%Begin of Solution. 

\begin{enumerate}[label={\arabic*.}]

\item  Proof of sufficiency. Suppose there is a function $U(x,y)$ in $\Omega$ with $\partial U/\partial x = p$ and $\partial U/\partial y= q$. 
\item  Let the curve $\gamma$ be defined by $z=z(t), a\le t\le b$. 
\item  We can compute the line integral explicitly, 
\begin{equation*}
\begin{aligned}
\int_\gamma p dx + q dy &= \int_\gamma \left( \frac{\partial U}{\partial x} dx + \frac{\partial U}{\partial y} dy  \right) 
 = \int_a^b \left( \frac{\partial U}{\partial x} x'(t) + \frac{\partial U}{\partial y} y'(t)  \right)dt 
 = \int_a^b \frac{dU}{dt}dt \\ 
&= U(x(b),y(b)) - U(x(a),y(a)). 
\end{aligned}
\end{equation*}
\ifnum\showsolution=2
\dotfill (\underline{\,\,4分\,\,})
\fi

\item  The integral depends only on the end points of $\gamma$. 

\item  Proof of necessity. We choose a fixed point $(x_0,y_0)$ in $\Omega$. 

\item  Since the line integral depends only on the end points of $\gamma$, we can define a function 
$$
U(x,y) = \int_\gamma pdx+qdy, 
$$
where $\gamma$ is any polygon contained in $\Omega$, connecting $(x_0,y_0)$ and $(x,y)$. 
\ifnum\showsolution=2
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\item  To compute $\frac{\partial U}{\partial x}$, we choose the last segment of $\gamma$ horizontal. 
Then 
$$
\frac{\partial U}{\partial x} = \lim\limits_{h\to 0} \frac{U(x+h,y)-U(x,y)}{h} 
= \lim\limits_{h\to 0} \frac{1}{h} \int_{(x-h,y)}^{x,y} p(x,y)dx = p(x,y). 
$$
\ifnum\showsolution=2
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\item  Similarly we can choose another path and verify that $\frac{\partial U}{\partial y}=q(x,y)$. 
\ifnum\showsolution=2
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\fi

\end{enumerate}

} % End of Solution. 

\fi

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%%第6题：解答结束

\vspace{0.2cm}

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%\newpage 
\item %7
Compute the complex line integral 
$$\int_{|z|=r} ydz$$ 
by observing that 
$ y=\frac{1}{2i}(z-\bar{z})=\frac{1}{2i}\left(z-\frac{r^2}{z}\right)$ on the circle. 

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{\color{red}解答： %Begin of Solution. 

\begin{enumerate}[label={\arabic*.}]
\item  Since the arc of the integral is the circle $|z|=r$, we see $z\bar{z}=r^2$ remains constant along the arc. 
\ifnum\showsolution=2
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\item  In the integral we can use the substitution, $$y=\frac{1}{2i}(z-\bar{z})=\frac{1}{2i}\left(z-\frac{r^2}{z}\right). $$
\ifnum\showsolution=2
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\item  Replace $y$ in the integral in terms of $z$ and we have 
\begin{equation*}
\begin{aligned}
\int_{|z|=r} ydz = \int_{|z|=r} \frac{1}{2i}\left(z-\frac{r^2}{z}\right)dz 
= \frac{1}{2i}\int_{|z|=r} zdz - \frac{r^2}{2i}\int_{|z|=r} \frac{dz}{z} = 0-\frac{r^2}{2i}\cdot 2\pi i = \pi r^2. 
\end{aligned}
\end{equation*}
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\end{enumerate}

} % End of Solution. 
\fi

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%%第7题：解答结束

\vspace{0.2cm}

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%\newpage 
\item %8
Let $\Omega$ be a region and $a\in\Omega$. 
Let $f(z)$ be an analytic function in $\Omega'=\Omega-\{a\}$. 
Prove that a necessary and sufficient condition that there exists an analytic function in $\Omega$ which coincides with $f(z)$ in $\Omega'$ is that $$\lim\limits_{z\to a} (z - a)f(z) = 0.$$ 

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\ifnum\showsolution>0

{\color{red}解答： %Begin of Solution. 

\begin{enumerate}[label={\arabic*.}]
\item  Necessity. Let $g(z)$ be an analytic function in $\Omega$ which coincides with $f(z)$ in $\Omega'=\Omega-\{a\}$. 

\ifnum\showsolution=2
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\item  Since the limit process does not involve the value of the function at the limiting point, we have 
$$\lim\limits_{z\to a} (z - a)f(z) = \lim\limits_{z\to a} (z - a)g(z) = 0\cdot g(a) =0. $$
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  Sufficiency. Consider some small circle $C=\{z : |z-a|=r \}\subset \Omega$. 

\item  Consider the complex integral with parameter $z$ inside $C$, 
$$g(z)=\frac{1}{2\pi i}\int_C \frac{f(\zeta)d\zeta}{\zeta - z}. $$
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\item  Use the condition $\lim\limits_{z\to a} (z - a)f(z) = 0$; by Cauchy's formula, we have $g(z)=f(z)$ for $0< |z-a|<r$. 

\item  The function $g(z)$ is analytic when $|z|<r$. 
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\end{enumerate}

} % End of Proof. 
\fi

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\vspace{0.2cm}

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%\newpage 
\item %9
Let $a>1$. Compute the definite integral 
$$
\int_0^{2\pi} \frac{d\theta}{a+\cos\theta}. 
$$

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{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  Let $z=e^{i\theta}$, and consider a complex integral along the unit circle $|z|=1$. 
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\item  Then $x=\cos\theta$, $y=\sin\theta$, and $dz=izd\theta$. 
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\item  We turn the real integral into a complex integral, 
$$
\int_0^{2\pi} \frac{d\theta}{a+\cos\theta}
= \int_{|z|=1} \frac{dz}{iz(a+x)} 
= \int_{|z|=1} \frac{dz}{iz[a+\frac{1}{2}(z+\frac{1}{z})] } 
= \int_{|z|=1} \frac{-2idz}{2az+z^2+1}.  
$$
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\item  Since $a>1$, we have the factorization $z^2+2a+1=(z-z_1)(z-z_2)$, where 
$$z_1=-a+\sqrt{a^2-1}, \,\, z_2=-a-\sqrt{a^2-1}. $$ 
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\item  Since $-1<z_1<0$ and $z_2<-1$, %by Cauchy's integral formula, 
\begin{equation*}
\begin{aligned}
\int_{|z|=1} \frac{-2idz}{2az+z^2+1}
&=\int_{|z|=1} \frac{-2idz}{(z-z_1)(z-z_2)}
=\int_{|z|=1} \frac{-2idz}{z_1-z_2}\left( \frac{1}{z-z_1} - \frac{1}{z-z_2} \right) \\ 
&=\int_{|z|=|} \frac{-2idz}{z_1-z_2} \frac{1}{z-z_1}  - \int_{|z|=1} \frac{-2idz}{z_1-z_2}\frac{1}{z-z_2} \\ 
&=2\pi i \frac{-2i}{z_1-z_2} - 0 
=\frac{4\pi}{z_1-z_2}
=\frac{2\pi}{\sqrt{a^2-1}}.
\end{aligned}
\end{equation*}
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\end{enumerate}

} % End of Proof. 
\fi


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\vspace{0.2cm}

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%\newpage 
\item %10
How many roots does the equation 
$z^7 + 2z^5 + 3z^3 + 11z + 4 = 0$ 
have in the disk $|z| < 1$? 
Hint: Look for the biggest term when $|z|=1$ and apply Rouche's theorem. 

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{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  Let $f(z)=z^7 + 2z^5 + 3z^3 + 11z + 4$ and $g(x)=11z$, then 
$f(z)-g(z)=z^7 + 2z^5 + 3z^3 + 4. $

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\item  When $|z|=1$, since 
\begin{equation*}
\begin{aligned}
|f(z)-g(z)| &= |z^7 + 2z^5 + 3z^3 + 4| \le 1+2+3+4=10, \\
|g(z)|=|11z| &= 11, 
\end{aligned}
\end{equation*}
we have $|f(z)-g(z)|<|g(z)|$. 
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\item  By Rouche's theorem, $f(z)$ and $g(z)$ have the same number of zeros in the disk. 
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\item  Since $g(z)$ has one root $z=0$ in the disk, so does $f(x)$. 
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\end{enumerate}

}
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\vspace{0.2cm}


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\end{enumerate}

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